graph the function ( y=\frac{2x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the…

graph the function ( y=\frac{2x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the derivatives ( y ) and ( y ), finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any. identify the absolute maximum value and where it occurs. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. a. the absolute maximum value ( square ) occurs at ( x=square ). (use a comma to separate answers as needed. type each answer only once ) b. there is no absolute maximum

graph the function ( y=\frac{2x}{x^{2}-16} ) by identifying the domain and any symmetries, finding the derivatives ( y ) and ( y ), finding the critical points and identifying the functions behavior at each one, finding where the curve is increasing and where it is decreasing, finding the points of inflection, determining the concavity of the curve, identifying any asymptotes, and plotting any key points such as intercepts, critical points, and inflection points. then find coordinates of absolute extreme points, if any. identify the absolute maximum value and where it occurs. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. a. the absolute maximum value ( square ) occurs at ( x=square ). (use a comma to separate answers as needed. type each answer only once ) b. there is no absolute maximum

Answer

Explanation:

Step1: Find the domain

The function (y = \frac{2x}{x^{2}-16}) is undefined when (x^{2}-16=0), i.e., (x = 4) or (x=- 4). So the domain is (\mathbb{R}-{-4,4}).

Step2: Check for symmetry

Replace (x) with (-x): (y(-x)=\frac{-2x}{x^{2}-16}=-y(x)). The function is odd, symmetric about the origin.

Step3: Find the first - derivative

Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 2x), (u^\prime=2), (v=x^{2}-16), (v^\prime = 2x). (y^\prime=\frac{2(x^{2}-16)-2x\times(2x)}{(x^{2}-16)^{2}}=\frac{2x^{2}-32 - 4x^{2}}{(x^{2}-16)^{2}}=\frac{-2x^{2}-32}{(x^{2}-16)^{2}}=\frac{-2(x^{2}+16)}{(x^{2}-16)^{2}}) Since (y^\prime=\frac{-2(x^{2}+16)}{(x^{2}-16)^{2}}<0) for all (x) in the domain (because (x^{2}+16>0) and ((x^{2}-16)^{2}>0) for (x\neq\pm4)), the function is decreasing on ((-\infty,-4)), ((-4,4)) and ((4,\infty)).

Step4: Analyze critical points

Set (y^\prime = 0), (\frac{-2(x^{2}+16)}{(x^{2}-16)^{2}} = 0). The numerator (-2(x^{2}+16)\neq0) for all real (x). So there are no critical points in the domain.

Step5: Analyze asymptotes

Vertical asymptotes: (x = 4) and (x=-4) (since the function is undefined at these points and (\lim_{x\rightarrow4^{-}}\frac{2x}{x^{2}-16}=-\infty), (\lim_{x\rightarrow4^{+}}\frac{2x}{x^{2}-16}=\infty), (\lim_{x\rightarrow - 4^{-}}\frac{2x}{x^{2}-16}=\infty), (\lim_{x\rightarrow - 4^{+}}\frac{2x}{x^{2}-16}=-\infty)) Horizontal asymptote: (\lim_{x\rightarrow\pm\infty}\frac{2x}{x^{2}-16}=\lim_{x\rightarrow\pm\infty}\frac{\frac{2x}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{16}{x^{2}}}=0)

Answer:

B. There is no absolute maximum.